Question: Simplify the following expression: $a = \dfrac{9z^2 + 63z + 54}{z + 1} $
Answer: First factor the polynomial in the numerator. We notice that all the terms in the numerator have a common factor of $9$ , so we can rewrite the expression: $ a =\dfrac{9(z^2 + 7z + 6)}{z + 1} $ Then we factor the remaining polynomial: $z^2 + {7}z + {6} $ ${1} + {6} = {7}$ ${1} \times {6} = {6}$ $ (z + {1}) (z + {6}) $ This gives us a factored expression: $\dfrac{9(z + {1}) (z + {6})}{z + 1}$ We can divide the numerator and denominator by $(z - 1)$ on condition that $z \neq -1$ Therefore $a = 9(z + 6); z \neq -1$